About the IODS project

I heard about this course in the MBDP Metagenomics course last spring. I wish to develop my R skills further and learn more about RMarkdown, version control and using GitHub.

1 Link to my GitHub repository

2 or: https://github.com/mnpuputti/IODS-project

IODS week 2. Regression and model validation: Data analysis

This week we practised data wrangling and data visualization for linear models in DataCamp with packages dplyr and ggplot2. We practised making scripts for data wrangling in R with “Learning2014” dataset. We used this dataset, modifed the data to match IODS-course needs, and we will call that part of the dataset “Students2014”. Dataset is available here.

Bringing data into R

Explore the “Students2014” dataset. This dataset contains part of the “the international survey of Approaches to Learning”

#Read data "students2014" into R from .csv-file created in the data wrangling part of week2.
students2014 <- read.csv("~/Documents/MAIJA/R_IODS/IODS-project/data/learning2014.csv", row.names = 1)
#Check dimensions and structure of the dataset.
dim(students2014)
## [1] 166   7
str(students2014)
## 'data.frame':    166 obs. of  7 variables:
##  $ gender  : Factor w/ 2 levels "F","M": 1 2 1 2 2 1 2 1 2 1 ...
##  $ age     : int  53 55 49 53 49 38 50 37 37 42 ...
##  $ attitude: int  37 31 25 35 37 38 35 29 38 21 ...
##  $ deep    : num  3.58 2.92 3.5 3.5 3.67 ...
##  $ stra    : num  3.38 2.75 3.62 3.12 3.62 ...
##  $ surf    : num  2.58 3.17 2.25 2.25 2.83 ...
##  $ points  : int  25 12 24 10 22 21 21 31 24 26 ...

There are 166 observations and seven different variables including: gender, age, attitude, deep, stra, surf and points. There are three different types of learning technique questions: “deep” = deep learning,“stra” = strategic learning and “surf” = surface learning. In addition there were categories for global attitude towards statistics “attitude” , exam points “points” and also the “gender” of the survey participants.

Load the libraries needed in data visualization

library(ggplot2)
library(dplyr)
library(GGally)

Check the data through visualizations

###Check graphical overview with basic plot of pairwise correlations
p1 <- plot(students2014, main = "Graphical overview of the students2014 dataset")

##Check graphical overview with finer, more informative visual presentation of pairwise correlation plot, with ggpairs
ggpairs(students2014, mapping = aes(col = gender, alpha=0.3), lower = list(combo = wrap("facethist", bins = 20)))

Summaries of the variables

###Check summaries of the different variables available 
summary(students2014)
##  gender       age           attitude          deep            stra      
##  F:110   Min.   :17.00   Min.   :14.00   Min.   :1.583   Min.   :1.250  
##  M: 56   1st Qu.:21.00   1st Qu.:26.00   1st Qu.:3.333   1st Qu.:2.625  
##          Median :22.00   Median :32.00   Median :3.667   Median :3.188  
##          Mean   :25.51   Mean   :31.43   Mean   :3.680   Mean   :3.121  
##          3rd Qu.:27.00   3rd Qu.:37.00   3rd Qu.:4.083   3rd Qu.:3.625  
##          Max.   :55.00   Max.   :50.00   Max.   :4.917   Max.   :5.000  
##       surf           points     
##  Min.   :1.583   Min.   : 7.00  
##  1st Qu.:2.417   1st Qu.:19.00  
##  Median :2.833   Median :23.00  
##  Mean   :2.787   Mean   :22.72  
##  3rd Qu.:3.167   3rd Qu.:27.75  
##  Max.   :4.333   Max.   :33.00
#or summaries could be also individually picked e.g.
summary(students2014$gender)
##   F   M 
## 110  56

From the graphical overview and summary we can see that there are more female participants in the gender category. Female participants are younger than male participants. Exam point means are on the similar level for both male and female participants. Male participants score better for the attitude points. There are no major correlations between any categories for either female or male participants.
Highest correlation with the “point” variable is observed with the variable “attitude” and “stra” and negative correlation *with “surf” variable“.* The mean age is 25.51 with range from 17.00 to 55.00 years. The mean observed exam points is 22.72 with minimum being 7.00 and maximum points being 33.00. Deep learning mean is 3.68, strategic learning mean is 3.121 and surface learning techniques mean is 2.787.

Regression model

#Create a regression model with three explanatory variables identified from pairwise correlation plot:stra + age + attitude,
model1 <- lm(points ~ stra + age + attitude, data = students2014)

#Check the summary of the regression model and plot the results
set.seed(123)
summary(model1)
## 
## Call:
## lm(formula = points ~ stra + age + attitude, data = students2014)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -18.1149  -3.2003   0.3303   3.4129  10.7599 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept) 10.89543    2.64834   4.114 6.17e-05 ***
## stra         1.00371    0.53434   1.878   0.0621 .  
## age         -0.08822    0.05302  -1.664   0.0981 .  
## attitude     0.34808    0.05622   6.191 4.72e-09 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 5.26 on 162 degrees of freedom
## Multiple R-squared:  0.2182, Adjusted R-squared:  0.2037 
## F-statistic: 15.07 on 3 and 162 DF,  p-value: 1.07e-08

Summary shows that only “attitude” of the tested explanatory variables could be considered statistically significant based on the p-values. Attitude has the most significant impact on the points with positive p.value 4.72e-09, while stra is showing p-value 0.0621 and age 0.0981. Attitude shows estimate of 0.34, while estimate for stra is 1.0. For age variable the estimate is negative. We can continue with the “attitude” variable and re-fit the model without the non-significant variables.

#Re-fit the model with significant variable "attitude"
model2 <- lm(points ~ attitude, data = students2014)

#Check the summary of the regression model and plot the results
set.seed(123)
summary(model2)
## 
## Call:
## lm(formula = points ~ attitude, data = students2014)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -16.9763  -3.2119   0.4339   4.1534  10.6645 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept) 11.63715    1.83035   6.358 1.95e-09 ***
## attitude     0.35255    0.05674   6.214 4.12e-09 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 5.32 on 164 degrees of freedom
## Multiple R-squared:  0.1906, Adjusted R-squared:  0.1856 
## F-statistic: 38.61 on 1 and 164 DF,  p-value: 4.119e-09

Now we can evaluate, whether the model and the data fit. The R-squared values tell us how close to the fitted (multiple) regression line our data is located. Multiple R-squared is 0.1906. This means that our model explains ~ 19 % of the variability around its mean. Estimates for the parametre for the model is 0.35255 for the attitude with standard error of 0.05674.

#Check how does the model looks with "points and "attitude"
qplot(attitude, points, data = students2014) + geom_smooth(method = "lm")

#Draw and explore the diagnostic plots
par(mfrow = c(2,2))
plot(model2, which = c(1,2,5))

Residuals are the errors in modelling the target variable. We want to minimize the modelling errors. Residuals vs fitted plot: There cannot be detected major spreading of the values. However, few outliers can be detected when fitted values increase. Normal QQ-plot explores the normal distribution potential of the errors. The data fits the normal distribution reasonably well, except the few outliers at the lower quantiles. We can say that the errors are normally distributed in this model. Residuals vs Leverage plot shows that there are no clear individual values that impact the model.

IODS Week 3: Logistic regression

This dataset contain data on student achievements in two Portuguese schools. We look at how alcohol consumption habits effects student performance.

Dataset student.zip-file can be found here. and information on dataset background here.

Let’s load needed libraries

library(dplyr)
library(ggplot2)
library(tidyr)

Reading in the data

Let’s read in the data from our data wrangling part. Alternatively, the modified dataset can be downloaded from here

alc <- read.csv("~/Documents/MAIJA/R_IODS/IODS-project/data/alc.csv", row.names = 1)

#Let's check the variables:
names(alc)  
##  [1] "school"     "sex"        "age"        "address"    "famsize"   
##  [6] "Pstatus"    "Medu"       "Fedu"       "Mjob"       "Fjob"      
## [11] "reason"     "nursery"    "internet"   "guardian"   "traveltime"
## [16] "studytime"  "failures"   "schoolsup"  "famsup"     "paid"      
## [21] "activities" "higher"     "romantic"   "famrel"     "freetime"  
## [26] "goout"      "Dalc"       "Walc"       "health"     "absences"  
## [31] "G1"         "G2"         "G3"         "alc_use"    "high_use"
#or use glimpse
glimpse(alc)
## Observations: 382
## Variables: 35
## $ school     <fct> GP, GP, GP, GP, GP, GP, GP, GP, GP, GP, GP, GP, GP, G…
## $ sex        <fct> F, F, F, F, F, M, M, F, M, M, F, F, M, M, M, F, F, F,…
## $ age        <int> 18, 17, 15, 15, 16, 16, 16, 17, 15, 15, 15, 15, 15, 1…
## $ address    <fct> U, U, U, U, U, U, U, U, U, U, U, U, U, U, U, U, U, U,…
## $ famsize    <fct> GT3, GT3, LE3, GT3, GT3, LE3, LE3, GT3, LE3, GT3, GT3…
## $ Pstatus    <fct> A, T, T, T, T, T, T, A, A, T, T, T, T, T, A, T, T, T,…
## $ Medu       <int> 4, 1, 1, 4, 3, 4, 2, 4, 3, 3, 4, 2, 4, 4, 2, 4, 4, 3,…
## $ Fedu       <int> 4, 1, 1, 2, 3, 3, 2, 4, 2, 4, 4, 1, 4, 3, 2, 4, 4, 3,…
## $ Mjob       <fct> at_home, at_home, at_home, health, other, services, o…
## $ Fjob       <fct> teacher, other, other, services, other, other, other,…
## $ reason     <fct> course, course, other, home, home, reputation, home, …
## $ nursery    <fct> yes, no, yes, yes, yes, yes, yes, yes, yes, yes, yes,…
## $ internet   <fct> no, yes, yes, yes, no, yes, yes, no, yes, yes, yes, y…
## $ guardian   <fct> mother, father, mother, mother, father, mother, mothe…
## $ traveltime <int> 2, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 1, 2, 1, 1, 1, 3,…
## $ studytime  <int> 2, 2, 2, 3, 2, 2, 2, 2, 2, 2, 2, 3, 1, 2, 3, 1, 3, 2,…
## $ failures   <int> 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,…
## $ schoolsup  <fct> yes, no, yes, no, no, no, no, yes, no, no, no, no, no…
## $ famsup     <fct> no, yes, no, yes, yes, yes, no, yes, yes, yes, yes, y…
## $ paid       <fct> no, no, yes, yes, yes, yes, no, no, yes, yes, yes, no…
## $ activities <fct> no, no, no, yes, no, yes, no, no, no, yes, no, yes, y…
## $ higher     <fct> yes, yes, yes, yes, yes, yes, yes, yes, yes, yes, yes…
## $ romantic   <fct> no, no, no, yes, no, no, no, no, no, no, no, no, no, …
## $ famrel     <int> 4, 5, 4, 3, 4, 5, 4, 4, 4, 5, 3, 5, 4, 5, 4, 4, 3, 5,…
## $ freetime   <int> 3, 3, 3, 2, 3, 4, 4, 1, 2, 5, 3, 2, 3, 4, 5, 4, 2, 3,…
## $ goout      <int> 4, 3, 2, 2, 2, 2, 4, 4, 2, 1, 3, 2, 3, 3, 2, 4, 3, 2,…
## $ Dalc       <int> 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,…
## $ Walc       <int> 1, 1, 3, 1, 2, 2, 1, 1, 1, 1, 2, 1, 3, 2, 1, 2, 2, 1,…
## $ health     <int> 3, 3, 3, 5, 5, 5, 3, 1, 1, 5, 2, 4, 5, 3, 3, 2, 2, 4,…
## $ absences   <int> 5, 3, 8, 1, 2, 8, 0, 4, 0, 0, 1, 2, 1, 1, 0, 5, 8, 3,…
## $ G1         <int> 2, 7, 10, 14, 8, 14, 12, 8, 16, 13, 12, 10, 13, 11, 1…
## $ G2         <int> 8, 8, 10, 14, 12, 14, 12, 9, 17, 14, 11, 12, 14, 11, …
## $ G3         <int> 8, 8, 11, 14, 12, 14, 12, 10, 18, 14, 12, 12, 13, 12,…
## $ alc_use    <dbl> 1.0, 1.0, 2.5, 1.0, 1.5, 1.5, 1.0, 1.0, 1.0, 1.0, 1.5…
## $ high_use   <lgl> FALSE, FALSE, TRUE, FALSE, FALSE, FALSE, FALSE, FALSE…

These parametres are fully explained at the data repository site. Key variables explained shortly: Dalc describes workday alcohol consumption, Walc weekend alcohol consumption, alc_use describes average alcohol use per day. All these three parametres have scale from 1 (very low) to 5 (very high use). High_use describes whether student uses alcohol more than twice a week.

Let’s see how the data looks in order to pick 4 interesting variables:

# draw a bar plot of each variable
gather(alc) %>% ggplot(aes(value)) + geom_bar(color="white",fill="pink") + facet_wrap("key", scales = "free")

Let’s pick variables: absences, school, age and sex and explore how they correlate with alcohol consumption. Our hypothesis is that males have higher rate in high use of alcohol than females. We can also hypothesize that high use of alcohol leads to more absences from school. We can test whether high alcohol use is linked to age. And we can see if there is a difference in culture towards alcohol use in these two different schools.

First, Let’s explore how students use alcohol in these two schools and whether sex can be associated with it.

plot1 <- ggplot(data = alc, aes(x=alc_use, fill=school))
plot1 + geom_bar() +
  facet_wrap("sex")

Here we can see that school “GP” shows higher average counts for alcohol use compared to school “MS”. In addition, the female participants have higher counts of very low or low average alcohol consumption compared to males.

Let’s see whether this changes for high_use

plot2 <- ggplot(data = alc, aes(x=high_use, fill=school))
plot2 + geom_bar() +
  facet_wrap("sex")

Also higher counts for high_alcohol consumption in school “GP” and male students. This confirms our hypothesis that male students consume more alcohol and that students in different schools may have different drinking culture and attitude towards drinking alcohol.

Let’s see what is the relationship with absences, high_alc usage and sex.

# plot1 explores the relationship of high_use and absences
plot3 <- ggplot(alc, aes(x = high_use, y = absences, col = sex))

# define the plot as a boxplot and draw it
plot3 + geom_boxplot() + ylab("absences") + ggtitle("Student absences by high alcohol consumption")

Here we can see that high use of alcohol increases the number of absences. Male students have more absences compared to female students. That is expected, since we already observed that male students have higher average alcohol usage.

# plot1 explores the relationship of high_use and absences with  variable age
plot4 <- ggplot(alc, aes(x = high_use, y = absences))

# define the plot as a boxplot and draw it
plot4 + geom_boxplot(color="darkblue") + ylab("absences") + facet_wrap("age") + ggtitle("Student absences by high alcohol consumption and age") + theme_bw()

Here we can see that high alcohol consumption is most common for 17 year old students and that they increase the absences.

Logistic regression

Let’s explore how high use of alcohol is linked to explanatory variables and use logistic regression

#We can now explore the relationship between four variables and high alcohol consumption with logistic regression
model1 <- glm(high_use ~ school+absences+sex+age, data = alc, family = "binomial")

#Let's study the summary of our fitted model
summary(model1)
## 
## Call:
## glm(formula = high_use ~ school + absences + sex + age, family = "binomial", 
##     data = alc)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -2.3081  -0.8357  -0.6384   1.0811   2.1077  
## 
## Coefficients:
##             Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -4.29534    1.79304  -2.396   0.0166 *  
## schoolMS     0.30068    0.39803   0.755   0.4500    
## absences     0.09467    0.02342   4.042 5.31e-05 ***
## sexM         0.99332    0.24134   4.116 3.86e-05 ***
## age          0.14593    0.10850   1.345   0.1786    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 465.68  on 381  degrees of freedom
## Residual deviance: 426.48  on 377  degrees of freedom
## AIC: 436.48
## 
## Number of Fisher Scoring iterations: 4

Here we can see that age and school are not statistically significant for high_use of alcohol despite our previous interpretation through visualization. However, we have factor variables here for school and sex and they are demonstrated in the intercept. Let’s explore the model without school variable.

#Let's re-fit our model, add -1 to exclude interception and show coefficient estimate values straight
model2 <- glm(high_use ~ sex+age+absences, data = alc, family = "binomial")

#Let's study the summary of our fitted model
summary(model2)
## 
## Call:
## glm(formula = high_use ~ sex + age + absences, family = "binomial", 
##     data = alc)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -2.3002  -0.8428  -0.6386   1.0781   2.1068  
## 
## Coefficients:
##             Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -4.73160    1.69730  -2.788  0.00531 ** 
## sexM         0.98440    0.24074   4.089 4.33e-05 ***
## age          0.17516    0.10125   1.730  0.08365 .  
## absences     0.09223    0.02314   3.986 6.71e-05 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 465.68  on 381  degrees of freedom
## Residual deviance: 427.04  on 378  degrees of freedom
## AIC: 435.04
## 
## Number of Fisher Scoring iterations: 4

Now we can see that coefficient for female students is seen in the intercept: -4.7 and for male student the estimate changed and is -0.98

#Let's study the coefficients of our model
coef(model2)
## (Intercept)        sexM         age    absences 
## -4.73159988  0.98439673  0.17515570  0.09223477
#Next we can explore the coefficients and their confidence intervals as odd ratios 
# We can compute odds ratios (oddrat)
oddrat <- coef(model2) %>% exp

# compute confidence intervals (confin)
confin <- confint(model2) %>% exp

#Npw we can print out our model's coefficients oddratios with their confidence intervals and join them
cbind(oddrat, confin)
##                  oddrat        2.5 %    97.5 %
## (Intercept) 0.008812361 0.0002990615 0.2356076
## sexM        2.676196933 1.6793970873 4.3232945
## age         1.191431709 0.9782468520 1.4562441
## absences    1.096622249 1.0500945691 1.1499088

Odd ratio describes the ratio of successes to failures. Here we can see that odds ratio for all explanatory variables is higher than 1 for all explanatory variables. This means that e.g. male students have 2.7 times probability to use high levels of alcohol than female students. Absences are likely to increase odds for high alcohol consumption.

Predictions

Absences and sex were statistically significant as explanatory variables. We can now re-fit our model with these variables and explore the predictions.

# re-fit the model with statistically significant variables
model3 <- glm(high_use ~ absences + sex, data = alc, family = "binomial")

# predict() the probability of high_use
probabilities <- predict(model3, type = "response")

# add the predicted probabilities to 'alc'
alc <- mutate(alc, probability = probabilities)

# use the probabilities to make a prediction of high_use
alc <- mutate(alc, prediction = probabilities>0.5)


# see the last ten original classes, predicted probabilities, and class predictions
select(alc, absences, sex, high_use, probability, prediction) %>%  tail(10) 
##     absences sex high_use probability prediction
## 373        0   M    FALSE   0.2976656      FALSE
## 374        7   M     TRUE   0.4545495      FALSE
## 375        1   F    FALSE   0.1493808      FALSE
## 376        6   F    FALSE   0.2215747      FALSE
## 377        2   F    FALSE   0.1620742      FALSE
## 378        2   F    FALSE   0.1620742      FALSE
## 379        2   F    FALSE   0.1620742      FALSE
## 380        3   F    FALSE   0.1756235      FALSE
## 381        4   M     TRUE   0.3841248      FALSE
## 382        2   M     TRUE   0.3395595      FALSE

Here we can see that all our predictions remain false and true cases of high_use are not predicted correctly. Let’s further explore this visually.

# tabulate the target variable versus the predictions
table(high_use = alc$high_use, prediction = alc$prediction)
##         prediction
## high_use FALSE TRUE
##    FALSE   258   10
##    TRUE     88   26

There were 88 false predictions of actual high_use and 26 accurate predictions of high_use.

# initialize a plot of 'high_use' versus 'probability' in 'alc'
g <- ggplot(alc, aes(x = probability, y = high_use, col = prediction))

# define the geom as points and draw the plot
g + geom_point()

Here we can see that the rate for true prediction increases, when probability reaches 0.5.

# tabulate the target variable versus the predictions
table(high_use = alc$high_use, prediction = alc$prediction) %>% prop.table %>% addmargins 
##         prediction
## high_use      FALSE       TRUE        Sum
##    FALSE 0.67539267 0.02617801 0.70157068
##    TRUE  0.23036649 0.06806283 0.29842932
##    Sum   0.90575916 0.09424084 1.00000000
Bonus: Exploration of cross-validation and wrong predictions
# the logistic regression model m and dataset alc (with predictions) are available

# define a loss function (average prediction error)
loss_func <- function(class, prob) {
  n_wrong <- abs(class - prob) > 0.5
  mean(n_wrong)
}

# compute the average number of wrong predictions in the (training) data


# K-fold cross-validation
library(boot)
crossval1 <- cv.glm(data = alc, cost = loss_func, glmfit = model3, K = 10)

# average number of wrong predictions in the cross validation
crossval1$delta[1]
## [1] 0.2670157

Since some of the non-significant variables were already excluded from our final model here, we explored the same variables as in DataCamp and model reached around the same 0.26 error.

Super-Bonus: exploration of different model options and their error rate
#Let's explore our previous models model 1 and model 2.
# K-fold cross-validation
#model 1
crossval2 <- cv.glm(data = alc, cost = loss_func, glmfit = model1, K = 10)

# average number of wrong predictions in the cross validation
crossval2$delta[1]
## [1] 0.2434555
# K-fold cross-validation
#model 2
crossval3 <- cv.glm(data = alc, cost = loss_func, glmfit = model2, K = 10)

# average number of wrong predictions in the cross validation
crossval3$delta[1]
## [1] 0.2486911
############# 
#And let's try a new test model to explore other variable options internet, school, age and failures
model4 <- glm(high_use ~ internet+school+age+failures, data = alc, family = "binomial")

# K-fold cross-validation
#model 4
crossval4 <- cv.glm(data = alc, cost = loss_func, glmfit = model4, K = 10)

# average number of wrong predictions in the cross validation
crossval4$delta[1]
## [1] 0.3062827
#Let's remove some of the variables one by one to see the effect on predictiosn
model5 <- glm(high_use ~ internet+age+failures, data = alc, family = "binomial")

# K-fold cross-validation
#model 4
crossval5 <- cv.glm(data = alc, cost = loss_func, glmfit = model4, K = 10)

# average number of wrong predictions in the cross validation
crossval5$delta[1]
## [1] 0.3036649
#And let's remove more variables:
model6 <- glm(high_use ~ internet+failures, data = alc, family = "binomial")

# K-fold cross-validation
#model 4
crossval6 <- cv.glm(data = alc, cost = loss_func, glmfit = model4, K = 10)

# average number of wrong predictions in the cross validation
crossval6$delta[1]
## [1] 0.3010471
#And let's try a new model to explore options:
model7 <- glm(high_use ~ internet, data = alc, family = "binomial")

# K-fold cross-validation
#model 4
crossval7 <- cv.glm(data = alc, cost = loss_func, glmfit = model4, K = 10)

# average number of wrong predictions in the cross validation
crossval7$delta[1]
## [1] 0.2984293
#And let's try a new model to explore options:
model7 <- glm(high_use ~ failures, data = alc, family = "binomial")

# K-fold cross-validation
#model 4
crossval7 <- cv.glm(data = alc, cost = loss_func, glmfit = model4, K = 10)

# average number of wrong predictions in the cross validation
crossval7$delta[1]
## [1] 0.3141361

Here we can see, that the test model 4 had increased prediction error (31 %) compared to previous model. Model 4 combined internet access, age, school and failures in our new test model. All the remaining attempt by removing one variable at the time did not decrease the prediction error.

IODS Week 4: Clustering and classification

Reading in the Boston dataset included in the MASS package and load in other needed libraries. Boston dataset contains Housing related data from Boston suburbs, and we explore the town crime rates with the explanatory variables from this dataset.

library(MASS)
library(tidyverse)
library(corrplot)
library(viridis)
library(ggplot2)
library(corrplot)
data(Boston)

Let’s explore the Boston dataset

str(Boston)
## 'data.frame':    506 obs. of  14 variables:
##  $ crim   : num  0.00632 0.02731 0.02729 0.03237 0.06905 ...
##  $ zn     : num  18 0 0 0 0 0 12.5 12.5 12.5 12.5 ...
##  $ indus  : num  2.31 7.07 7.07 2.18 2.18 2.18 7.87 7.87 7.87 7.87 ...
##  $ chas   : int  0 0 0 0 0 0 0 0 0 0 ...
##  $ nox    : num  0.538 0.469 0.469 0.458 0.458 0.458 0.524 0.524 0.524 0.524 ...
##  $ rm     : num  6.58 6.42 7.18 7 7.15 ...
##  $ age    : num  65.2 78.9 61.1 45.8 54.2 58.7 66.6 96.1 100 85.9 ...
##  $ dis    : num  4.09 4.97 4.97 6.06 6.06 ...
##  $ rad    : int  1 2 2 3 3 3 5 5 5 5 ...
##  $ tax    : num  296 242 242 222 222 222 311 311 311 311 ...
##  $ ptratio: num  15.3 17.8 17.8 18.7 18.7 18.7 15.2 15.2 15.2 15.2 ...
##  $ black  : num  397 397 393 395 397 ...
##  $ lstat  : num  4.98 9.14 4.03 2.94 5.33 ...
##  $ medv   : num  24 21.6 34.7 33.4 36.2 28.7 22.9 27.1 16.5 18.9 ...
dim (Boston)
## [1] 506  14

Boston dataset has 506 observations and 14 different variables. More information on different variables is available here.

#Let's explore the Boston data with graphical overview and summaries of the variables
pairs(Boston)

summary(Boston)
##       crim                zn             indus            chas        
##  Min.   : 0.00632   Min.   :  0.00   Min.   : 0.46   Min.   :0.00000  
##  1st Qu.: 0.08204   1st Qu.:  0.00   1st Qu.: 5.19   1st Qu.:0.00000  
##  Median : 0.25651   Median :  0.00   Median : 9.69   Median :0.00000  
##  Mean   : 3.61352   Mean   : 11.36   Mean   :11.14   Mean   :0.06917  
##  3rd Qu.: 3.67708   3rd Qu.: 12.50   3rd Qu.:18.10   3rd Qu.:0.00000  
##  Max.   :88.97620   Max.   :100.00   Max.   :27.74   Max.   :1.00000  
##       nox               rm             age              dis        
##  Min.   :0.3850   Min.   :3.561   Min.   :  2.90   Min.   : 1.130  
##  1st Qu.:0.4490   1st Qu.:5.886   1st Qu.: 45.02   1st Qu.: 2.100  
##  Median :0.5380   Median :6.208   Median : 77.50   Median : 3.207  
##  Mean   :0.5547   Mean   :6.285   Mean   : 68.57   Mean   : 3.795  
##  3rd Qu.:0.6240   3rd Qu.:6.623   3rd Qu.: 94.08   3rd Qu.: 5.188  
##  Max.   :0.8710   Max.   :8.780   Max.   :100.00   Max.   :12.127  
##       rad              tax           ptratio          black       
##  Min.   : 1.000   Min.   :187.0   Min.   :12.60   Min.   :  0.32  
##  1st Qu.: 4.000   1st Qu.:279.0   1st Qu.:17.40   1st Qu.:375.38  
##  Median : 5.000   Median :330.0   Median :19.05   Median :391.44  
##  Mean   : 9.549   Mean   :408.2   Mean   :18.46   Mean   :356.67  
##  3rd Qu.:24.000   3rd Qu.:666.0   3rd Qu.:20.20   3rd Qu.:396.23  
##  Max.   :24.000   Max.   :711.0   Max.   :22.00   Max.   :396.90  
##      lstat            medv      
##  Min.   : 1.73   Min.   : 5.00  
##  1st Qu.: 6.95   1st Qu.:17.02  
##  Median :11.36   Median :21.20  
##  Mean   :12.65   Mean   :22.53  
##  3rd Qu.:16.95   3rd Qu.:25.00  
##  Max.   :37.97   Max.   :50.00

From the summary we can see e.g. the mean crime rate in Boston suburbs is 3.6 crimes/capita. nox tells us the nitrogen oxide concentrations (ppm) with mean value in Boston of 0.6. dis tells us that the distance from the Boston employment centres ranges from 1.1 to 12.

Correlation plot
#or exploration with corrplot
# First we need to calculate the correlation matrix, optionally print it and round it
cor_matrix<-cor(Boston) 
#cor_matrix
roudedcormatrix<- cor_matrix %>% round(digits = 2) 

#plot 
corrplot(roudedcormatrix, type = "upper", cl.pos = "b", tl.pos = "d", tl.cex = 0.6)

From the correlation plot we can see the positive and negative correlations with different variables and crime rate (crim). Crime rate has the highest positive correlation with variables rad (index of accessibility to radial highways) and tax (full-value property-tax rate per $10,000). On the opposite, lowest correlations are detected with medv (median value of owner-occupied homes in $1000s), black (1000(Bk - 0.63)^2 where Bk is the proportion of blacks by town) and dis (weighted mean of distances to five Boston employment centres).

Scaling
#Let's standardise our dataset and scale our data
Bostonscaled <- scale(Boston)

#Let's change our matrix into a dataframe 
Bostonscaled<-as.data.frame(Bostonscaled)

#Check the class 
class(Bostonscaled)
## [1] "data.frame"
#Let's check the summary
summary(Bostonscaled)
##       crim                 zn               indus        
##  Min.   :-0.419367   Min.   :-0.48724   Min.   :-1.5563  
##  1st Qu.:-0.410563   1st Qu.:-0.48724   1st Qu.:-0.8668  
##  Median :-0.390280   Median :-0.48724   Median :-0.2109  
##  Mean   : 0.000000   Mean   : 0.00000   Mean   : 0.0000  
##  3rd Qu.: 0.007389   3rd Qu.: 0.04872   3rd Qu.: 1.0150  
##  Max.   : 9.924110   Max.   : 3.80047   Max.   : 2.4202  
##       chas              nox                rm               age         
##  Min.   :-0.2723   Min.   :-1.4644   Min.   :-3.8764   Min.   :-2.3331  
##  1st Qu.:-0.2723   1st Qu.:-0.9121   1st Qu.:-0.5681   1st Qu.:-0.8366  
##  Median :-0.2723   Median :-0.1441   Median :-0.1084   Median : 0.3171  
##  Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.:-0.2723   3rd Qu.: 0.5981   3rd Qu.: 0.4823   3rd Qu.: 0.9059  
##  Max.   : 3.6648   Max.   : 2.7296   Max.   : 3.5515   Max.   : 1.1164  
##       dis               rad               tax             ptratio       
##  Min.   :-1.2658   Min.   :-0.9819   Min.   :-1.3127   Min.   :-2.7047  
##  1st Qu.:-0.8049   1st Qu.:-0.6373   1st Qu.:-0.7668   1st Qu.:-0.4876  
##  Median :-0.2790   Median :-0.5225   Median :-0.4642   Median : 0.2746  
##  Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.: 0.6617   3rd Qu.: 1.6596   3rd Qu.: 1.5294   3rd Qu.: 0.8058  
##  Max.   : 3.9566   Max.   : 1.6596   Max.   : 1.7964   Max.   : 1.6372  
##      black             lstat              medv        
##  Min.   :-3.9033   Min.   :-1.5296   Min.   :-1.9063  
##  1st Qu.: 0.2049   1st Qu.:-0.7986   1st Qu.:-0.5989  
##  Median : 0.3808   Median :-0.1811   Median :-0.1449  
##  Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.: 0.4332   3rd Qu.: 0.6024   3rd Qu.: 0.2683  
##  Max.   : 0.4406   Max.   : 3.5453   Max.   : 2.9865
#Let's create quantiles for the crime rates 
crimequantiles<- quantile(Bostonscaled$crim)
crimequantiles
##           0%          25%          50%          75%         100% 
## -0.419366929 -0.410563278 -0.390280295  0.007389247  9.924109610
#and rename them with labelstring
labelstring<-c("low","med_low","med_high","high")
crime <- cut(Bostonscaled$crim, breaks = crimequantiles, include.lowest = TRUE, label = labelstring)

#Let's make changes into the dataset and remove original crim
Bostonscaled <- dplyr::select(Bostonscaled, -crim)

# add the new categorical value to scaled data
Bostonscaled <- data.frame(Bostonscaled, crime)
Now we can divide the dataset into test and train sets
#Let's see the number of rows in our dataset
n <- nrow(Bostonscaled)

#We can choose randomly 80% of the rows
ind <- sample(n,  size = n * 0.8)

# We can create train set
train <- Bostonscaled[ind,]

# We can create test set 
test <- Bostonscaled[-ind,]

Linear discriminant analysis (LDA)

#We can now use the categorical crime rate as the target variable and all the other variables in the dataset as predictor variables (as dot in script)
lda.fit <- lda(crime ~.,data = train)

lda.fit
## Call:
## lda(crime ~ ., data = train)
## 
## Prior probabilities of groups:
##       low   med_low  med_high      high 
## 0.2698020 0.2202970 0.2574257 0.2524752 
## 
## Group means:
##                   zn      indus        chas        nox          rm
## low       0.94776710 -0.8971760 -0.09172814 -0.8661508  0.41403330
## med_low  -0.09773272 -0.2625464  0.08156758 -0.5620961 -0.13125022
## med_high -0.41220520  0.2703592  0.25766519  0.4240806 -0.00487157
## high     -0.48724019  1.0171096 -0.04073494  1.0406599 -0.38642231
##                 age        dis        rad        tax     ptratio
## low      -0.8731314  0.8574965 -0.6984422 -0.7588152 -0.40831234
## med_low  -0.3444234  0.3681568 -0.5534543 -0.4905469 -0.06872784
## med_high  0.4573932 -0.4064705 -0.3999076 -0.2478149 -0.20286305
## high      0.8080500 -0.8529113  1.6382099  1.5141140  0.78087177
##                black       lstat         medv
## low       0.38256024 -0.76416040  0.499383470
## med_low   0.35237238 -0.11838895  0.009138486
## med_high  0.08931233  0.08857418  0.056652597
## high     -0.84222907  0.83994701 -0.633454131
## 
## Coefficients of linear discriminants:
##                 LD1         LD2         LD3
## zn       0.10049915  0.72763600 -0.93452410
## indus   -0.06232910 -0.25446786  0.46592588
## chas    -0.06856708 -0.09745969  0.11177269
## nox      0.35900138 -0.52823345 -1.38164587
## rm      -0.10933789 -0.06468908 -0.08375578
## age      0.32618623 -0.34883089 -0.24539175
## dis     -0.11348351 -0.22009469  0.31728047
## rad      2.91204021  1.07347198  0.18122290
## tax      0.11480221 -0.22548808  0.37179438
## ptratio  0.10464710  0.05358570 -0.31555558
## black   -0.15215701 -0.01281596  0.17254010
## lstat    0.16466416 -0.22434855  0.48464256
## medv     0.14840845 -0.29710857 -0.12316117
## 
## Proportion of trace:
##   LD1   LD2   LD3 
## 0.944 0.043 0.013

From the summary of LDA we can see that LDA model explains over 90 % of the variance. In addition, we can see the different explanatory variable means, and we can check how they vary between crime rates, e.g. age, nox and dis show variation between low crime and high crime rate means.

LDA biplot
# Let's create the function for lda biplot arrows
lda.arrows <- function(x, myscale = 1, arrow_heads = 0.1, color = "color", tex = 0.75, choices = c(1,2)){
  heads <- coef(x)
  arrows(x0 = 0, y0 = 0, 
         x1 = myscale * heads[,choices[1]], 
         y1 = myscale * heads[,choices[2]], col=color, length = arrow_heads)
  text(myscale * heads[,choices], labels = row.names(heads), 
       cex = tex, col=color, pos=3)
}

# Set target classes as numeric
classes <- as.numeric(train$crime)

# Now we can biplot the LDA results
p <- plot(lda.fit, dimen = 2, col = classes, pch = classes) 

p + lda.arrows(lda.fit, col = classes, myscale = 1.5) 

## integer(0)

Or We can also try ggplot2 and viridis packages for LDA biplot visualization Example script can be found here and more on data management prior to ggplotting can be found here

require(scales)
require(gridExtra)

#We can also try ggplot2 and viridis packages for LDA biplot visualization 

r <- lda(formula = crime ~ ., 
         data = train, 
         prior = c(1,1,1,1)/4)

prop = r$svd^2/sum(r$svd^2)

##
lda <- lda(crime ~ .,data=train, prior = c(1,1,1,1)/4)

prop.lda = r$svd^2/sum(r$svd^2)

plda <- predict(object = lda, newdata = train)


dataset2 = data.frame(crime = train[,"crime"], lda = plda$x)

p1 <- ggplot(dataset2) + geom_point(aes(lda.LD1, lda.LD2, colour = crime), size = 2.5) + 
  labs(x = paste("LD1 (", percent(prop.lda[1]), ")", sep=""),
y = paste("LD2 (", percent(prop.lda[2]), ")", sep="")) +
    scale_color_viridis(discrete = TRUE, option = "plasma") +
  theme_bw()

p1

We can also try ordinations in ggord instead.

#library(devtools)
#install_github("fawda123/ggord")
library(ggord)

ord <- lda(crime ~ ., train, color = crime, prior = rep(1, 4)/4)
ggord(ord, train$crime, color = crime, arrow=0.6, txt=4, size=3)

From the plots we can see that only group clearly separating from other groups is the high crime rate group. In addition, we can see that the explanatory variable rad points out towards high crime rate and nox points out towards med_high crime rate, where as explanatory variable zn points out towards low crime rate. This means that rad, which was “index of accessibility to radial highways”, could almost solely separate the high crime rate grouping in this model.

library(dplyr)

# We can now save the new classes from test data
correct_classes <- test$crime
correct_classes
##   [1] low      med_low  med_low  med_high med_high med_high med_high
##   [8] med_low  med_low  med_low  med_low  med_low  med_low  med_low 
##  [15] med_low  low      med_low  low      low      med_low  med_low 
##  [22] med_low  med_low  med_low  med_low  med_low  med_low  med_low 
##  [29] med_low  med_low  med_high med_high med_high med_high med_high
##  [36] med_low  low      low      low      med_low  low      low     
##  [43] med_low  low      low      low      med_low  med_high med_high
##  [50] med_high med_high med_high med_high med_low  med_low  med_high
##  [57] med_high med_high med_high med_high med_high med_low  low     
##  [64] med_low  med_low  low      med_low  med_high med_low  low     
##  [71] low      low      low      med_low  high     high     high    
##  [78] high     high     high     high     high     high     high    
##  [85] high     high     high     high     high     high     high    
##  [92] high     high     high     high     high     high     high    
##  [99] high     med_low  med_low  med_low 
## Levels: low med_low med_high high
# And we can remove the crime variable from test data
test <- dplyr::select(test, -crime)

Predictions

#Predictions of classes with LDA model and test data
set.seed(123)
lda.pred <- predict(lda.fit, newdata = test)

# And we can then cross tabulate the results
predictions<-table(correct = correct_classes, predicted = lda.pred$class)
predictions
##           predicted
## correct    low med_low med_high high
##   low       12       5        1    0
##   med_low    7      18       12    0
##   med_high   1       7       14    0
##   high       0       0        0   25

Predictions with the test data show that 14 of low crime rate was correctly predicted as low and 5 falsely as med_low. In addition, 5 med_high crime rated was falsely predicted as low and 9 as med_high. From the high crime rate 31 was correctly predicted as high. The model seems to work well with the high crime rate predictions. However, there can be seen more false predictions for med_high, med_low and low crime rates.

NOTE! Despite the set.seed(), this table seems to change the values with every knit and I did not find a solution to this yet

Predicting no. of clusters

#We can reload the Boston dataset
data(Boston)

#Let's standardise the dataset by scaling the variables, 
#we need data.frame
Bostonscaled2 <- as.data.frame(Boston)

#Now we can use k-means clustering to calculate the distances
# k-means clustering
km <-kmeans(Bostonscaled2, centers = 4)
#km$cluster

#plot
pairs(Bostonscaled2[1:10], col = km$cluster)

####
set.seed(123)
# determine the number of clusters
k_max <- 10

# calculate the total within sum of squares
twcss <- sapply(1:k_max, function(k){kmeans(Bostonscaled2, k)$tot.withinss})

# visualize the results
qplot(x = 1:k_max, y = twcss, geom = 'line')

From the qplot we can see that the optimal number of clusters is 2, since there is a radical rise.

#Let's set the number of centers to 2 and cluster again.

# k-means clustering
km2 <-kmeans(Bostonscaled2, centers = 2)

# plot the Boston dataset with clusters
pairs(Bostonscaled2[1:10], col = km2$cluster)

When we compare the clustering of crime vs explanatory variables, good grouping can be assumed e.g. in rad and tax. In comparison, nox and rm scatter more.

#Bonus

#Superbonus